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If $s_n$ is an increasing sequence of positive numbers, define $\sigma_n := \frac{1}{n}(s_1 + s_2 + ... + s_n)$. Show that if $s_n \to L$, then $\sigma_n \to L$ as well.

The hint that was given stated:

$\forall \epsilon > 0$, choose $N_0$, then $\forall n > N_0, |s_n - L| < \epsilon $. Consider $\sigma_n$, split the sum into two parts - $n \leq N_0$ and $n > N_0$. The later can be bounded trivially. For the former, note that the sum over the prefix $n \leq N_0$ is bounded (by what?) and show that for $n$ large enough (depending on the $N_0$), this sum is negligble.

I'm having trouble understanding what "split the sum into two parts" means in this context.

Thanks for any help.

Jimmy Katz
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  • Sum the stuff below $N_0$, then the stuff above $N_0$. – Randall Feb 05 '23 at 17:48
  • We are given that $s_n$ is an increasing sequence and $s_n \to L$. Here's a similar reasoning: Choose $N_0$ such that $\forall n> N_0, |s_n - L|<\epsilon$ for some very small $\epsilon$. Since $N_0$ is finite, $\sigma_{n_0}$ is finite and hence does not influence $\sigma_n$ as $n \to \infty$ – D S Feb 05 '23 at 17:52
  • Thank you guys. I understand it now. – Jimmy Katz Feb 05 '23 at 18:24

1 Answers1

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When it says "split the sum into two parts: $n\le N_0$ and $n>N_0$", they are saying to split it like this $$\frac{s_1+s_2+s_3+...+s_{N_0}}{n}+\frac{s_{N_0+1}+s_{N_0+2}+...s_n}{n}$$

Kamal Saleh
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